
L O.
asked 01/12/14help in solving by completing the square
I am trying to solve 2x^2 + 3x =20 by completing the square.
I rewrote it to be 2x^2 - 3x + 20 = 0
then started with 2x^2/2 + 3x/2 + 20/10 = x^2 - 3/2x + 10 = 0
how do I continue?
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3 Answers By Expert Tutors

Steve S. answered 01/12/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
"I am trying to solve 2x^2 + 3x =20 by completing the square."
2x^2 + 3x = 20
Divide both sides by 2:
x^2 + (3/2)x = 10
We want to force a pattern that gives us a perfect square; i.e.,
(x-h)^2 = x^2 - 2hx + h^2.
So force our equation into having that pattern:
x^2 - 2(-3/4)x + (-3/4)^2 = 10 + (-3/4)^2
Replace the perfect square trinomial on the left
with a perfect square:
(x + 3/4)^2 = 10 + 9/16 = 169/16 = (13/4)^2
Take the square root of both sides (√(z^2) = |z|):
|x + 3/4| = 13/4
Solve for x:
x + 3/4 = ±13/4
x = -3/4 ± 13/4
x = (-3 ± 13)/4
x = -4 or 5/2

Parviz F. answered 01/12/14
Tutor
4.8
(4)
Mathematics professor at Community Colleges
2X^2 -3X + 20 =0
2 ( X^2 -3/ 2 X + 10 ) = 0
( X ^2 - 2( 3/4 ) X ) = -10
( X^2 - 2(3/4) X + 9/16) = -10 + 9/16
( X - 3/ 4 ) ^2 = -151/ 16
X - 3/4 = ± √151/4 i
X = 3/4 ± i√151/4
There are several ways to solve a quadratic equation by completing the square.
The approach I like best is to work with the equation in the form you had originally:
2 x2 +3 x = 20 {the constant term on the right hand side}
From this we identify a as a = 2 and b as b= 3. The number to add to both sides is b2/4a = 9/8
This results in
2 x2 + 3 x + 9/8 = 20 + 9/8 = 169/8
Next we divide the equation by 2 to get the coefficient of x2 to be 1
x2 + (3/2) x + 9/16 = 169/16
Because we added the b2/4a , the left hand side will be perfect square. It is easy to see that it is the square of (x + 3/4) Thus we have
(x + 3/4)^2 = 169/16
We next take the square root of both sides to get
(x + 3/4) = ± 13/ 4
So finally x = -3/4 ± 13/4 or
x = 5/2 and x = -4
With this general approach, the only thing to memorize is that the quantity to add is b2/4a
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Vivian L.
01/12/14